Answer
The statement is true.
Work Step by Step
Let us consider the left hand side expression of the provided equation, $_{7}{{P}_{3}}$
Recall, $_{n}{{\operatorname{P}}_{r}}=\frac{n!}{\left( n-r \right)!}$
Therefore, $\begin{align}
& _{7}{{P}_{3}}=\frac{7!}{\left( 7-3 \right)!} \\
& =\frac{7!}{4!} \\
& =\frac{7\times 6\times 5\times 4!}{4!} \\
& =210
\end{align}$
Also, consider the left hand side expression of the provided equation, $3!\cdot \left( _{7}{{C}_{3}} \right)$
Recall, $_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$
Then, $\begin{align}
& 3!\cdot \left( _{7}{{C}_{3}} \right)=3!\times \frac{7!}{3!\left( 7-3 \right)!} \\
& =3!\times \frac{7!}{3!\left( 4 \right)!} \\
& =\frac{7!}{\left( 4 \right)!} \\
& =\frac{7\times 6\times 5\times 4!}{4!}
\end{align}$
So, $3!\cdot \left( _{7}{{C}_{3}} \right)=210$
Thus, $_{7}{{P}_{3}}=3!\cdot \left( _{7}{{C}_{3}} \right)$
